Precalculus

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Luciano Battaia, Giacomo Bormetti, Giulia Livieri

A Prelude to Calculus with Exercises

Notes for a crash-course in Mathematics, Università di Bologna,

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Luciano Battaia, Giacomo Bormetti, Giulia Livieri Version 1.0 of November 13, 2019

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The power of mathematics is often to change one thing into another, to change geometry into language. Marcus du Sautoy

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A Word from the Authors

In this work you will find all the contents of a crash-course in Mathematics, intended for students at the first year of a University Course in Economics.

If you find any mistake, please let us know at the mail addressbatmath@gmail.com.

Luciano Battaia, Università Ca’ Foscari di Venezia, Dipartimento di Economia Giacomo Bormetti, Università di Bologna, Dipartimento di Matematica

Giulia Livieri, Scuola Normale Superiore di Pisa and Università di Bologna, Scuola di Economia, Management e Statistica

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1 Basic concepts

The aim of this chapter is to recall some basic mathematical concepts that will be needed in later chapters: special products and factors, radicals and fractions.

1.1 Special products and factors

Some mathematical problems require the computation of products involving two or more variables to simplify expressions and to obtain the desired results. Many of these products arespecial because they are very common, and they are worth knowing. If we are able to recognize these products easily, it makes our life easier later on. On the other hand, it is also important to be able to write an algebraic sum as the product of its simplest factors, i.e. performingfactorization or factoring.

In what follows, some (of the most common) examples of special products and factors are presented. 1.1.1 Factoring an algebraic expression

Consider the following examples.

Example 1.1. 6x+ 2x2y+ 4xy2= 2x(3 + xy + 2y2). This means that the factors of 6x + 2x2y+ 4xy2are the monomial 2x and the polynomial 3+ xy + 2y2_{, and that the}_{common factor is 2x.}

Example 1.2. a2_b_{+ ab}2_{= ab(a + b).}

Example 1.3. (a + b)2_{− 2b (a + b ) + 2a(a + b ) = (a + b )(a + b − 2b + 2a) = (a + b )(3a − b ). Here the} common factor(a + b) is a polynomial and not a monomial as in previous Examples1.1and1.2. Example 1.4. 3b2(x2+ y) − 6b3(x2+ y) + 12b4(x2+ y) = 3b2(x2+ y)(1 − 2b + 4b2). The common factors are the monomial 3b2and the polynomial(x2+ y).

Sometimes, factoring is more hard-working

Example 1.5. ax+ ay + b x + b y = a(x + y) + b(x + y) = (x + y)(a + b), and, sometimes, more creative

Example 1.6. ax− b x + b y − ay − b + a = x(a − b ) − y(a − b ) + (a − b ) = (a − b )(x − y + 1). In algebra, multiplying binomials is easier if we are able to recognize their patterns. We multiply sum and difference of binomials and multiply by squaring and cubing to find some of the special products. 1.1.2 Product of a sum and a difference

If we multiply the sum and the difference of two quantitiesa and b we get the following rule

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So, the product of a sum and a difference of the same two terms is equal to the difference of the squares of the terms. If we read Equation (1.1) from right to left we see that the difference of squares of two terms is equal to the product of the sum and the difference of the same two terms.

The quantitiesa and b can be any two monomials and polynomials.

Note 1.1. Note that it is not possible to recognize a pattern for the sum of squares of two terms a2_{+ b}2_. Consider now the following examples.

Example 1.7. (x − 1)(x + 1) = x2− 1.

Example 1.8. (−x − 2)(−x + 2) = (−x)2_{− 4 = x}2_{− 4.} Example 1.9. x2− 3 = (x −p3)(x +p3).

Example 1.10. (xy + 2x + 3y)(xy + 2x − 3y) = [(xy + 2x) + 3y][(xy + 2x) − 3y] = (xy + 2x)2− (3y)2= x2_y2_{+ 4x}2_y_{+ 4x}2_{− 9y}2_.

1.1.3 Square of a binomial

To compute(a + b)2, i.e. square of a binomial, we have the following rule

(1.2) (a + b)2= a2+ 2ab + b2.

So, to square(a + b), we square the first term (a2), add twice the product of the two terms (2ab), then add the square of the last term(b2). Similarly, to compute (a − b)2, we have the following rule

(1.3) (a − b)2= a2− 2ab + b2.

So, to square(a − b), we square the first term (a2), subtract twice the product of the two terms (−2ab), then add the square of the last term(b2_).

The quantitiesa and b can be any two monomials and polynomials. Equations (1.2) and (1.3) permit also to factor the special trinomialsa2+ 2ab + b2anda2− 2ab + b2.

Consider the following examples.

Example 1.11. (x + 2y)2= x2+ 2 · x · 2y + (2y)2= x2+ 4xy + 4y2.

Example 1.12. (4x − 3y)2= (4x)2− 2(4x)(3y) + (3y)2= 16x2− 24xy + 9y2. Example 1.13. x2_{+ 4x + 4 = (x + 2)}2_.

The techniques introduced so far can also be combined together as showed in the following examples. Example 1.14. x3+ 6x2+ 9x = x(x2+ 6x + 9) = x(x + 3)2.

Example 1.15. (x + y − z)(x + y + z) = [(x + y) − z][(x + y) + z] = (x + y)2− z2= x2+ 2xy + y2− z2. Example 1.16. (x +y + z)2_{= [(x +y)+ z]}2_{= (x +y)}2_{+2(x +y)z + z}2_{= x}2_{+2xy +y}2_{+2xz +2yz + z}2_. So, the square of a sum of an arbitrary number of terms becomes:

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Precalculus 1.2 Radicals

1.1.4 Cube of a binomial

For cubing the sum and the difference of two quantitiesa and b , we have the following two rules (1.4) (a + b)3= a3+ 3a2b+ 3ab2+ b3, (a − b)3= a3− 3a2b+ 3ab2− b3.

So, the cube of a sum (difference) of two quantitiesa and b is equal to the cube of the first term, plus (minus) three times the square of the first term by the second term, plus (plus) three times the first term by the square of the second term, plus (minus) the cube of the second term.

Example 1.17. (2x + y)3= (2x)3+ 3(2x)2y+ 3 · 2x(y)2+ y3= 8x3+ 12x2y+ 6xy2+ y3. Example 1.18. (x2− 3y)3= (x2)3− 3(x2)23y+ 3x2(3y)2− (3y)3= x6− 9x4y+ x2y2− 27y3. Example 1.19. a3b3− 3a2b2+ 3ab − 1 = (ab − 1)3.

1.1.5 Sum and difference of cubes

Contrary to squares,both sum and difference of two cubes can be decomposed. The following rules hold

(1.5) a3+ b3= (a + b)(a2− ab + b2), a3− b3= (a − b)(a2+ ab + b2). Note 1.2. The middle of the trinomials is always opposite the sign of the binomial.

Note 1.3. The two trinomials a2_{− ab + b}2_and_a2_{+ ab + b}2_{are not squares because the double product} is not present.

We consider some examples.

Example 1.20. (x3_{− 1) = (x − 1)(x}2_{+ x + 1).}

Example 1.21. (8x3+ 27y3) = (2x + 3y)(4x2− 6xy + 9y2).

1.2 Radicals

In many situations, it is useful to simplify mathematical expressions involving radicals, without trying to rewrite them using decimal approximations. The following example on numerical approximations clarifies the importance of this statement. Suppose we have to compute(p2)8. Applying only the exponent properties, we obtain(p2)8_{= (}p₂₎24

= 24_{= 16. On the other hand, if we first approximate} p

2≈ 1.4, and then we raise this approximation to the power of 8 we obtain (p2)8≈ 1.4)8= 14.76. So, the error that we make is not negligible!

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The main properties of radicals are listed below n p an_{= a ,} pn _an = a (definition); n p ab= pn _apn _b _{(product property);} n È_a b = n p a n p b (quotient property); (pn am = pn am _{(power property);} n p

an_bp_{= a}pn _bp _{(factor out n-powers);}

n pp

am p₌ pn

am _{(simplify rational exponents).} (1.6)

In previous expressions, it is assumed thata and b are positive real numbers when n and p are even integers(1).

Note 1.4. There is no property linked to the n-t h root of a sum of two quantities a and b : pn a+ b 6=

n

p

a+ pn b .

Note 1.5. It is possible to sum two radicals only if they are similar. It is possible to multiply two radicals only if they have the same index.

Example 1.22. 5p8+ 3p2= 5p22₂+ 3p₂= 5 · 2p₂+ 3p₂= 13p_2. Example 1.23. 3p27−p12+p2= 3p32₃₋p₂2₃+p₂= 3 · 3p₃_{− 2}p₃+p₂= 7p₃+p_2. Example 1.24. p2p3 2=p6 23p6 ₂2=p6 8· 4 =p6 32.

1.3 Algebraic fractions

An algebraic function is theratio between two polynomials. For example, x3+ xy + y2+ 2

x2_{− y}

is an algebraic fraction. The methodology used to simplify algebraic fractions is exactly the same used to simplify numerical fractions. Consider the following examples.

Example 1.25. x 2_{+ x} x2_{− 1} + x+ 2 x− 1= x(x + 1) (x − 1)(x + 1) +x+ 2 x− 1= x+ x + 2 x− 1 = 2x+ 2 x− 1 . Example 1.26. 3x(x + 2) x+ 1 · x− 1 x+ 2= 3x(x + 2) x+ 1 · x− 1 x+ 2 =3x(x − 1) x+ 1 = 3x2− 3x x+ 1 . Example 1.27. x 2_{− 1} x3_{+ 1}= (x − 1)(x + 1) (x + 1)(x 2_{− x + 1)}= x− 1 x2_{− x + 1}.

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2 Sets, Numbers and Functions

The aim of this chapter is to review certain mathematical concepts and tools which should be known to the reader. In particular, the following concepts are presented: i) the summation symbol, ii) the general concept of set along with various relations and operations, iii) numbers and, in particular, intervals of real numbers, iv) functions. In what follows the symbol N indicates the set of natural numbers, Z the set of integer numbers, Q the set of rational numbers and, finally, R the set of real numbers.

2.1 The summation symbol

The summation symbolΣ (capital sigma) was introduced around 1820 by the physicist and mathe-matician J. Fourier (1768-1830). It is a very convenient way to write complicated formulae.

Suppose that we want to write the sum of the integer numbers 1, 2, 3. In this case, we can write 1+2+3. However, if we want to write the sum of the integer numbers from 1 to 100(1)we (probably) write

(2.1) 1+ 2 + ··· + 99 + 100,

where the dots warn that the summation involves also the numbers from 3 to 98, not explicitly displayed. To represent (2.1) more parsimoniously, we can write

100 X

i=1 i

which readsthe sum of i for i going from 1 to 100. In general, however, the addends of (2.1) can be more complex. For example, they can be:

– the reciprocal of the natural numbers:1/_i_, – the square of the natural numbers: i2,

– any expression involving the natural numbers, such as the ratio between a natural number and its consecutive.

– etc.

Generally speaking, given a finite sequence of terms

(2.2) a₁,a₂,· · · , an

1_{It is reported that at the age of 7 the Prince of Mathematicians Friedrich Gauss (1777-1855) amazed his teacher by summing}

the integers from 1 to 100 almost instantly, having quickly spotted that the sum was actually 50 pairs of numbers, with each pair summing to 101, total 5 050.

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(which readsa sub one, a sub two, etc.), to denote their sum we can choose a letter, say i, as an index ranging fromm to n, and write

n X

i=m a_i,

which readsthe sum of a (sub) i for i (going) from m to n. a_iis called thegeneral term. The value of the sumdoes not depend on the name chosen for the index, but only on its range. For this reason, we say that the summation index is adummy index.

In particular the sums

n X i=m a_i , n X j=m a_j e n X k=m a_k are equivalent. The following examples will fix the concepts:

– 10 X i=5 1 i2 = 1 52+ 1 62+ 1 72+ 1 82+ 1 92+ 1 102; – 100 X i=2 i i− 1= 2 2− 1+ 3 3− 1+ ··· + 99 99− 1+ 100 100− 1; – 5 X i=0 (−1)i_{= (−1)}0_{+ (−1)}1_{+ (−1)}2_{+ (−1)}3_{+ (−1)}4_{+ (−1)}5_{= 1 − 1 + 1 − 1 + 1 − 1 (= 0) .} The summation symbol is subject (intuitively) to the same properties of the sum operation. In particular, theassociative property holds

n X k=1 a_k= m X k=1 a_k+ n X k=m+1 a_k (m < n). The following examples conclude this section.

Examples. – 100 X i=2 2i+ 4 i− 1 = 2 100 X i=2 i+ 2 i− 1; – 20 X i=0 (−1)i i = (−1) 20 X i=0 (−1)i−1 i .

2.2 Sets

A set is identified by explicitly declaring theobjects (the elements) belonging to it, or a property which characterises them. Sets are usually denoted by capital letters such asA, B, . . . , while their elements are denoted by small lettersa, b , . . . .

A typical symbol in Set theory, which is too important to be given up, is the symbol∈, which indicates that an elementbelongs to a set. Writing

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Precalculus 2.2 Sets

means that the elementx belongs to the set A. To say, instead, that the element x does not belong to the setA the writing

(2.4) x /∈ A or A 63 x

is used. Two sets areequal if and only if they have the same elements. Using the symbol∀ (for all), we have the following statement

(2.5) A= B ⇔ (∀x x ∈ A ⇔ x ∈ B),

where the symbol⇔ stands for if and only if. In particular, the ordering with which the elements are listed is not relevant, but only the elements themselves matter.

Among the various sets there is a very special one, without any elements, called theempty set and denoted by;. From Equation (2.5) it follows that the empty set is unique.

In general, two representations are used in order to describe a set:

1. Extensive representation: all the elements of a set are explicitly listed between curly brackets. For instance,A=¦0,π,p2, Pordenone, Forlí©.

2. Intensive representation: all the elements of a set are implicitly listed through a common property. For instance,A= { x | x is an even natural number }.

Verifying whether an elementx belongs to a set A is not a trivial task. Suppose, for example, A= P, the set of prime numbers. While it is immediate to verify that 31∈ P , it is more difficult to check that also 15 485 863∈ P . However, to verify that 243 112 609− 1 ∈ P(2)a long computation time is required, even using powerful computers.

We say thatA is a subset of B, or that A is contained (or included) in B, or that B is a superset of A, if every element inA is also an element of B. We write

A⊆ B , B⊇ A.

The inclusion symbol,⊆, does not exclude the possibility that A and B coincide. If we want to rule out this possibility, the symbol ofproper (or strict) inclusion must be used, i.e.,

A⊂ B B⊃ A,

which readsA is strictly included in B. In particular, the empty set; is strictly included in every other set. IfA6= ; and A ⊂ B, we also say that A is a proper subset of B. Every set A has as improper subsets A itself and;. In this course, we are also interested in sets with only one element: if a ∈ A, then { a } ⊆ A. Note 2.1. Symbols∈ and ⊂ have a very different meaning. The first one links two different objects (an element and a set), while the second links objects of the same type (two sets).

Finally, given a setA, the set of all subsets of A is given the name of power set and is denoted byP (A). For instance, ifA= { a, b }, then

P (A) = { ;, { a } , { b } , A} .

2_{This is one of the biggest prime numbers known at the end of 2009, with 12 978 189 digits. We would like to stress that most}

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2.3 Relations and operations with sets

Definition 2.1 (Union of two sets). The union of two sets A and B, denoted by A∪ B, is the set of elements x such that x belongs to A, to B, or both(3)

A∪ Bdef= { x | x ∈ A∨ x ∈ B } .

Example 2.1. If A= { 0, 1, 2, 3 } and B = { 2, 3, 4 }, then A∪ B = { 0, 1, 2, 3, 4 }.

Definition 2.2 (Intersection of two sets). The intersection of two sets A and B, denoted by A∩ B, is the set of the elements x such that x belongs to A and to B

A∩ Bdef= { x | x ∈ A∧ x ∈ B } . Example 2.2. Let A and B be as in Example2.1, thenA∩ B = { 2, 3 }.

If two sets have an empty intersection, i.e., ifA∩ B = ;, they are called disjoint. The empty set ; is disjoint with itself and with all sets.

The operations introduced above show strong analogies with the arithmetic operations, where the union plays the role of the sum, whereas the intersection plays the role of the product. In particular, the associative property, which is immediately verifiable and where A, B and C denote any three sets, holds true

(A∪ B) ∪ C = A∪ (B ∪ C ), (A∩ B) ∩ C = A∩ (B ∩ C ). (as a consequence, it is possible to write simplyA∪ B ∪ C and A∩ B ∩ C ). Besides, the following relations hold

A∪ A = A; A∩ A = A; A∪ B = B ∪ A; A∩ B = B ∩ A;

A∪ ; = A; A∩ ; = ;; A∪ B ⊇ A; A∩ B ⊆ A;

A∪ B = A ⇔ A ⊇ B; A∩ B = A ⇔ A ⊆ B. Thedistributive property reads as

A∪ (B ∩ C ) = (A∪ B) ∩ (A∪ C ) , A∩ (B ∪ C ) = (A∩ B) ∪ (A∩ C ) .

Note 2.2. There are, actually, two distributive properties. One of the union with respect to the intersec-tion, and another one of the intersection with respect to the union. Instead, in the arithmetic operations case, only the distributive property of the product with respect to the sum is valid:a(b + c) = ab + ac. Definition 2.3 (Difference between sets). Given two sets A and B, the difference between A and B, denoted by A\ B or A− B, is the set made up of the elements which belong to A but not to B.

A\ Bdef= { x | x ∈ A∧ x /∈ B } .

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Precalculus 2.4 Numbers

Example 2.3. Let A and B as in Example2.1, thenA\ B = { 0, 1 }.

In particular, ifB⊆ A, the set A\ B is named complementary set of B with respect to A. If the role of A is clear it will be denoted by ûAB or ûB. Often, it happens that all of the sets under consideration are subsets of a common setU , called the universe set. In this case, we simply write ûB instead of ûUB.

The Set theory presented so far is the so callednaive theory. Although sufficient for our purposes, it presents some problems: in particular, paradoxes can arise(4).

We have seen that the sets{a, b } and {b , a} coincide because they have the same elements. In many practical situations, however, it is important to deal withordered pairs, where the ordering in which the elements are written does matter. More precisely, given two setsA and B, an ordered pair, denoted by (a, b), is obtained by choosing an element a ∈ A and an element b ∈ B in the specified order. In symbols

(a, b) = (a0,b0) ⇔ a = a0∧ b = b0. It is convenient to observe that, in general:

{ a, b }= { b,a } while (a, b) 6= (b,a).

Definition 2.4 (Cartesian product). Given two sets A and B, the Cartesian product, or simply product of A and B, denoted by A× B, is the set of all the ordered pairs (a, b ), with a ∈ A and b ∈ B. In formulae

A× Bdef= { (a, b) | (a ∈ A) ∧ (b ∈ B) } .

Given the importance of the ordering in the pair(a, b), it should be clear that, whenever A is different fromB, A× B 6= B × A. In the case when A = B, you write A× A = A2.

Finally, you can also consider Cartesian products of more than two sets and, in the case of the Cartesian product of a set with itselfn times you write An= A× A× ··· × A

| {z }

n times .

2.4 Numbers

The building blocks of mathematics are numbers. In particular, the following sets of numbers are frequently used

N, Z, Q, R .

– N is the set of natural numbers. The mathematician Leopold Kronecker (1823-1891) used to say thatnatural numbers are God’s creation. In what follows, the set of natural numbers is

N = { 0, 1, 2, . . . , n, . . . } .

This set has a minimum element, the 0, but not a maximum element. Precisely, every subset of the set of natural numbers admits a minimum element.

4_{Maybe the most famous one is the barber paradox due to Bertrand Russel (1872-1970). The paradox is the following:}_The

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– Z (the symbol derives from the German word zahl, which means number, digit) is the set of integer numbers. Broadly speaking, integer numbers are natural numbers with sign, with the exception of the 0 (+0 = −0 = 0)

Z = { . . . , −2, −1, 0, 1, 2, . . . } . Each natural and integer number admits aconsecutive.

– Q (the symbol is due to the fact that a rational number is essentially a quoziente - Italian for quo-tient) is the set ofrational number, i.e., numbers which can be represented as ratios (or fractions) of integer numbers, where care is taken not to put the number 0 at the denominator.

Q = {m/n| m ∈ Z, n ∈ N, n 6= 0 } .

There are infinitely manyequivalent fractions representing the same rational number. For instance, 1/7 is equivalent to 2/14, to 3/21, and so on. Among them, it is particularly convenient to consider fractionsreduced to its lowest terms, where the numerator and the denominator are prime with respect to each other.

Rational numbers admit also a decimal representation, illustrated in the following examples. To represent 2/5 you get 2/5 = 0.4. But dividing 214 by 495, instead, you obtain 214/495 = 0.4323232.... In the first case, afinite amount of numbers is required after the decimal point, whereas, in the second case, the digits 32 repeat indefinitely. It is said that the alignment is periodical and that 32 is the period. Differently from the previous two sets of numbers, you cannot speak about theconsecutive of a rational number. In particular, between two rational numbers there is an infinite number of rational numbers:

if a= m

n and b= p

q than the number c= a+ b

2 is a rational number betweena and b . – R is the set of real numbers. In this course, we do not want to describe rigorously this set of

numbers. Broadly speaking, the set of real numbers can be thought as the set of all integer numbers, fractions, radicals, the numbers asπ, etc.

The following relations hold

N ⊂ Z ⊂ Q ⊂ R .

Common to all of these sets, there is the possibility to do sums and products. However, it is not always possible to perform subtractions in N and divisions in Z. Sometimes, it may happen that we have to use the set of complex numbers. This set is denoted by C and it is a superset of R. The main advantage is that, within the set of complex numbers, it is always possible to compute the square root of a negative number.

2.5 Intervals of real numbers

Some subsets of R, which are called intervals, deserve special attention: In this section, we give the definition and we consider some properties of these subsets.

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Precalculus 2.6 Functions

Definition 2.5. Given two real numbers a and b , with a< b, you call intervals the following subsets of R

]a,a[= (a,a) ; empty interval

]a, b[= (a, b) { x | a< x < b } bounded interval open

[a, b] { x | a ≤ x ≤ b } bounded interval closed

[a, b[= [a, b) { x | a ≤ x < b } bounded intervalleft-closed and right-open ]a, b] = (a, b] { x | a< x ≤ b } intervallo limitato left-open and right-closed ]a,+∞[= (a,+∞) { x | x> a } left-bounded and right-unbounded left-open [a,+∞[= [a,+∞) { x | x ≥ a } left-bounded and right-unbounded left-closed ] − ∞,a[= (−∞,a) { x | x< a } left-unbounded and right-bounded right-open ] − ∞,a] = (−∞,a] { x | x ≤ a } left-unbounded and right bounded right-closed The numbers a and b are the extremes of the interval. Bounded intervals are also named segments, whereas unbounded intervals half lines.

Consistently, we can write R =] − ∞, +∞[ or, equivalently, R = (−∞, +∞), which is the only unbounded interval whose geometrical image is the entire straight line. It is both open and closed. Whena= b the closed interval [a,a] has only one element and it is named degenerate interval. Sometimes, also the empty set; is considered as an interval. In particular, it is named empty interval.

If[a, b] is a bounded interval, i.e. [a, b]def= {x ∈ R : a ≤ x ≤ b} the point x₀= a+ b

2 is namedcenter while the number

δ = b − x0= x0− a

radius. In particular, an open interval with center x₀and radiusδ is given by ]x0− δ , x0+ δ[.

Each point of an interval that does not coincide with the (possible) extremes is named interior point.

2.6 Functions

In applications, relations between two setsA and B have great interest. Among these relations, functions between two intervals of real numbers deserve particular attention. The following definition holds.

Definition 2.6. Given two sets A and B, a function from A to B is a law that associates with each element of A one (single) element of B. The set A is called the domain of the function, while the set B is called the co-domain. If x is an element of the set A and y is the unique element of the set B corresponding to A, it is said that y is function of x; in symbols y= f (x).

It is important to remind that, in order to assign a function it is necessary to specify – the domain

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– the co-domain

– a law that associates with the elementx in the domain the unique element y in the co-domain. Different notations are used to indicate a function. The most complete one is the following

f : A→ B, x 7→ f (x) , but, often, the symbol

x7→ f (x) ,

is used if the setsA and B have already been defined or their definition is clear from the context. However, the most used one is the less rigorous notationy= f (x) (to be read y is f of x).

Example 2.4. If A and B are the set of real numbers, we can consider the function that associates with the real numberx∈ A the element y = x2_in_{B. So, we can use one of the following three notations}

f : R → R, x 7→ x2, or

x7→ x2 or

y= x2.

Often, to visualize a function arrow diagrams are used. For instance

b x4 b x1 b x2 b x3 b y₁ b y₂ b y₃ b y₄ b y₅ A B

Figure 2.1 Arrow diagram to visualize a function between two finite sets

In particular, itis required that from each point of A originates exactly one arrow. Conversely, it can happen that a point ofB is “shot” with more than one arrow, or that it is not “shot” at all.

In applications, the so-calledimage set has a particular interest. It is given by

(2.6) I ⊆ Bdef= { y ∈ B | ∃x ∈ A, y = f (x) },

and defined as the set of all possibleoutputs y∈ B coming from all possible inputs x ∈ A. The set I is denoted by the symbol f(A). If C is a subset of A, then we can consider the set f (C ) ⊆ f (A).

Other types of representations can be used for functions. For instance, let f be the function that associates with the elementx in A= {1,2,3,4,5} the element y in B = {1/2,1,3/2,2,5/2} (note that the

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Precalculus 2.6 Functions

domain of f is the subset of natural numbers{1, 2, 3, 4, 5}, whereas the co-domain is the set of rational numbers). The following tabular representation can be used

x x/₂ 1 1/₂ 2 1 3 3/₂ 4 2 5 5/₂ .

Table 2.1 Tabular representation of a function

In particular, in the first column there are the natural numbers 1, 2, . . . , 5 whereas in the second the corresponding halves.

Another type of representation is thepie chart. For instance, let us consider an undergraduate program where 120 students, coming from different provinces, are enrolled:

Gorizia Pordenone Treviso Trieste Udine

5 70 15 10 20.

To build the pie-chart, first we compute the percentages relative to each province

4.17 58.33 12.5 8.33 16.67 ,

then we map these percentages into the slices of the pie chart, taking into account that the whole pie measures 360◦_:

15◦ ₂₁₀◦ ₄₅◦ ₃₀◦ ₆₀◦

At this point the chart is immediate

4.17 % Gorizia (5) 58.33 % Pordenone (70) 12.5 % Treviso (15) 8.33 % Trieste (10) 16.67 % Udine (20)

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Finally, a bar-chart it is also used

5

70

15 ₁₀ 20

Figure 2.3 Bar-chart indicating the origin of 120 students enrolled in an undergraduate course

As mentioned, we are mainly interested in numerical functions, whereinput and output variables are numbers or groups of numbers. In particular, in this course real numbers come into play as variables, and for this reason we will talk aboutreal functions of a real variable. In all these cases, we are dealing with laws associating with a real numberx one real number y only, so that they have a subset A of R as domain and R as co-domain.

In order to visualize the behaviour of a function, the study of itsgraph in an appropriate Cartesian plane turns out to be very useful. In particular, the following definition holds

Definition 2.7. The graph of a function f : A→ B is the set of pairs (x, y), with x ∈ A, y ∈ B, such that y= f (x).

For instance, if we consider example in Table2.1, we have to represent the points A= (1,1/₂), B = (2,1), C = (3,3/₂), D = (4,2), E = (5,5/₂), into the followinggraph:

1 2 3 1 2 3 4 5 6 −1 b A b B b C b D b E

Figure 2.4 Cartesian graph

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Precalculus 2.6 Functions 0.5 1.0 1.5 2.0 2.5 3.0 −0.5 1 2 3 4 5 6 −1 A B C D E

Figure 2.5 Cartesian graph, with arrows.

Representing a function into a Cartesian graph has many advantages. This is clear if we compare Figure2.4with Table2.1. In particular, from the graph it is immediate to figure out that the function ismonotonic (precisely, it is a monotonically increasing function) and that its growth is steady. The advantages are more evident if we consider the function that associates with the real numberx∈ R the elementy= x/2 in R. In this case, the variable x assumes an infinite number of values and so it is not possible to have a tabular representation(5). In particular, we have the following Cartesian graph:

0.5 1.0 1.5 2.0 −0.5 −1.0 −1.5 −2.0 1 2 3 4 −1 −2 −3 −4

Figure 2.6 Cartesian graph of the function y=x/₂

Obviously, graph in Figure2.7contains also the points represented in the graph of Figure2.4:

0.5 1.0 1.5 2.0 2.5 3.0 −0.5 −1.0 −1.5 1 2 3 4 5 6 −1 −2 b A Bb b C Db b E

Figure 2.7 Cartesian graph of the function y=x/₂_{. Some points are put in evidence.}

5_{Note that, however, the law that associates with the element}_{x the element y is the same of the previous case: In order to}

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Often it is required to study a real function of a real variable in order to draw an indicative graph of it. Some software are devoted to this end(6). However, we have to keep in mind that computers are not build up to solve all our problems! For instance, let consider the graph of the following function

f(x) = sin1/_x_. 1 2 3 4 5 −1 −2 −3 −4 −5

Figure 2.8 Graph of f(x) = sin1/_x

It should be clear that asx approaches 0, the graph in Figure2.8is not very significant. So, we try to zoom the horizontal axis:

1 −1

Figure 2.9 Graph of f(x) = sin1/_x_{. We zoom the horizontal axis.}

Figure 2.10 Graph of f(x) = sin1/_x_{. We zoom the horizontal axis of the graph in Figure}_2.9_.

without success. Le us now consider a luckier example. Precisely, the graph of the function f(x) = x3−3x2is obtained, with sufficient accuracy, with a commercial software and displayed in the following figure.

6_{Among the commercial software, we point out two sophisticated, although complicated, packages:}_{Mathematica and Maple.}

Instead,Maxima (a less sophisticated version of Mathematica) and Geogebra are two non commercial software. In particular,

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Precalculus 2.7 Exercises 1 −1 −2 −3 −4 1 2 3 −1 −2 −3 Figure 2.11 Grafico di f(x) = x3_{− 3x}2

Observing the graph, it is evident that the function f(x) = x3− 3x2is: i) strictly increasing on the interval] − ∞,0], ii) strictly decreasing on the interval [0,2] and, iii) strictly increasing on the interval [2,+∞[.

All the graphs, except those in Figure2.9and2.10, present the same unit of measure on both axes. Cartesian systems of this type are namedmono-metric. However, in practice, it is not always possible to use mono-metric Cartesian systems. For instance, both graphs in Figure below represent a circumference center at the origin and with unitary radius. However, only the left panel employs a mono-metric Cartesian system. −1 1 1 −1 −1 1 1 −1

Figure 2.12 Circumference with center at the origin and unitary radius. The left panel employs a mono-metric Cartesian system. The right one, instead, has not a Cartesian system and it appears warped.

2.7 Exercises

Exercice 2.1. Given the sets A=] − ∞,2], B = { 1,2 } e C = [0,5[, determine 1. (A\ C ) ∪ B;

2. (A\ B) ∪ C ; 3. (A\ B) ∩ C ; 4. (C \ B) ∩ A.

Exercice 2.2. Given the sets A= { 1 }, B =] − 1,2[ e C =]0,+∞[, determine 1. (A∪ C ) ∩ B;

2. A\ C ; 3. (C \ A) ∩ B; 4. (C ∪ B) \ A; 5. (b \ A) ∩ C .

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Exercice 2.3. Discuss succinctly, yet unequivocally, the following questions. 1. Is it possible to find three sets A, B, C such that(A∩ B) ∪ C = ;? 2. Is it possible to find three sets A e B such that A∩ B = A?

3. Is it possible to find three sets A, B, C such that(A∩ B) ∪ C = A? 4. If A⊆ B then (C \ B) ⊆ (C \ A).

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3 Equations

3.1 Linear equations

In this section we look at linear equations in one variablex.

The most generallinear equation – this means there will be no x2_{terms and no}_x3_{’s, just}_{x’s and numbers} – inone variable is of the type

(3.1) ax= b , a 6= 0.

Equation in (3.1) admits a unique solution(1)

(3.2) x= b

a . Ifa∈ R, then three cases must be considered:

– a6= 0: the equation has only one solution x =b/_a_; – a= 0 ∧ b 6= 0: the equation has no solution ;

– a= 0 ∧ b = 0: the equation has an infinite number of solutions (basically all R).

In particular, it is important to consider the above conditions when solving parametric equations. For instance, consider the following example.

Example 3.1. Solve the following equation:

(a2_{− 1)x = a + 1 .} To solve it, we have to take into account the following cases:

– Ifa6= ±1, than it has only one solution x =(a + 1)/(a2_{− 1)}=₁/_{(a − 1)}_;

– Ifa= 1, than it has no solution;

– Ifa= −1 , than it has an infinite number of real solutions. The most generallinear equation in two variables is of the type

(3.3) ax+ b y = c , (a, b) 6= (0,0).

The condition on the parametersa and b is equivalent to say that they are not both zero at the same time. Equation (3.3) has always an infinite number of solutions. To obtain these solutions one first

1_The_{Fundamental Theorem of Algebra states that an equation of grade n has at maximum n solutions in R. As a consequence}

equation (3.1) has always one solution. This is not the case if we consider non linear equations. For this type of equations

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transforms equation (3.3) into a linear equation in one variable by fixing one of the two variables, then solves the latter as previously discussed. For instance, the following equation

2x+ 3y = 1 has as solutions the pairs(0,1/₃), (1/₂, 0), (−1,1), etc.

The following one variable equation, can be considered as a two variables equation where the coeffi-cient ofy is 0:

3x= 1, or 3x + 0y = 1, and it has as solutions the pairs(1/₃, 1), (1/₃, 2), (1/₃_,_{−5), etc.}

3.2 Two dimensional systems of linear equations

In mathematics, asystem of linear equations in two variables is a collection of two linear equations involving the same set of variables. For example

(3.4)

ax+ b y = p

c x+ d y = q,

is a system of linear equations in the two variablesx and y. The grade of the system is obtained as product of the grades of each equation. In particular, the system in Equation (3.4) has grade equal to one.

The wordsystem indicates that equations have to be considered collectively, rather than individually. Asolution to a linear system is an assignment of numbers to the variables such that all the equations are simultaneously satisfied. In particular, the system is said

– determinate if it has a unique solution;

– indeterminate if it has an infinite number of solutions; – inconsistent if it has no solution.

In general, a system of linear equations isconsistent if there is at least one set of values for the unknowns that satisfies every equation in the system.

Consider the following examples: –

2x+ y = 1

x− y = 2 : The system isconsistent and determinate. It has as unique solution the pair (1,−1).

–

x− 2y = 1

2x− 4y = 2 : The system isconsistent and indeterminate. It has as solutions the pairs(2t + 1,t)∀t ∈ R.

–

x− 2y = 1

2x− 4y = 3 : The system isinconsistent.

One method of solving a system of linear equations in two variables is theby substitution method. The method of solvingby substitution works by solving one of the two equations (we choose which one) for one of the variables (we choose which one), and then plugging this back into the other equation,

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Precalculus 3.3 Second order equations

substituting for the chosen variable and solving for the other. Then we back-solve for the first variable. For instance 2x+ y = 1 x− y = 2 , y= 1 − 2x x− y = 2 , y= 1 − 2x x− (1 − 2x) = 2 , y= 1 − 2x x= 1 , y= −1 x= 1 .

3.3 Second order equations

This section is aboutsingle-variable quadratic equations and their solutions. The most generalquadratic equation is any equation having the form

(3.5) ax2+ b x + c = 0, a 6= 0,

wherex represents an unknown, and a, b , and c represent known numbers such that a is not equal to 0. Ifa= 0, then the equation is linear, not quadratic. A quadratic equation can be solved using the general quadratic formula (3.6) x_1,2= −b ± p b2_{− 4ac} 2a . In particular, it admits

– two distinct solutions if the quantity∆ = b2−4ac (named discriminant or simply Delta) is greater than zero;

– one solution (it is said that the quadratic equation hastwo coincident real solutions or that it has a double solution) if∆ = 0;

– no solution in R if ∆ < 0. In this case there are two solutions in the complex set C. To fix ideas, we consider the following examples.

Examples. – 2x2_{− 3x − 5 = 0} _{=⇒ x} 1,2= 3±p9 − 4 · 2(−5) 2· 2 = 3±p49 4 = 5/₂ −1 ; – x2_{− 6x + 9 = 0} _{=⇒ x} 1,2= 6±p36− 4 · 9 2 = 3;

– x2− 2x + 2 = 0 =⇒ there is no solution because ∆ = 4 − 4 · 2 < 0.

3.4 Higher order equations

There are general formulae for solving cubic (third degree polynomials) and quartic (fourth degree polynomials) equations. However, we are not interested to them in this course. Instead, there are no formulae to solve general equations having grade greater than 4. We limit our analysis to two simple cases.

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3.4.1 Elementary equations

An elementary equation is an equation of type

(3.7) axn+ b = 0, a 6= 0.

In order to solve Equation (3.7), we have to make the following steps: i) “take” the termb to the other side of the equation (while changing the sign), ii) divide the latter term bya, iii) find the n-th root of the term−b /a. It is important to make attention if n is an odd or an even number.

Example 3.2. 2x3+ 54 = 0 =⇒ x3= −27 =⇒ x = −3.

Example 3.3. 3x3− 12 = 0 =⇒ x3_{= 4 =⇒ x =}p3

4.

Example 3.4. 2x4_{+ 15 = 0 =⇒ x}4_{= −}15/₂ =⇒ There is no solution. Example 3.5. 3x4− 14 = 0 =⇒ x4=14/₃ =⇒ x = ±p4

14/₃_.

3.4.2 Factorizable equations We only consider some example.

Example 3.6. x3− x2= 0 =⇒ x2(x − 1) = 0 =⇒ x2= 0 ∨ x − 1 = 0 =⇒ x = 0 ∨ x = 1. Example 3.7. x3−1 = 0 =⇒ (x−1)(x2+x+1) = 0 =⇒ x = 1 only, as the equation x2+x+1 = 0 has no solution (∆ < 0).

Example 3.8. x4−1 = 0 =⇒ (x2−1)(x2+1) = 0 =⇒ (x−1)(x+1)(x2+1) = 0 =⇒ x = ±1. (The equationx2_{+ 1 = 0 has no solution).}

3.5 Equations with radicals

A “radical” equation is an equation in which at least one variable expression is stuck inside a radical (in this course we consider the case of square roots or cubic roots).

There is no standard technique to solve radical equations. In general, we solve equations by isolating the variable. So, in the radical equation case we first have to isolate the square (or cubic) root, then to square (or to cube) both members. The new equation does not contain any radical. It is important to remind that, in the square root case, we have to check if all the solutions of the new equation are consistent with the initial one. There are no checks to do in the cubic root case. For instance

Example 3.9. px+ 2 + x = 0, px+ 2 = −x, x + 2 = x2, x2− x − 2 = 0, x_1,2= 1± p 1+ 8 2 = −1

2 , only the solutionx= 1 is consistent.

Example 3.10. px+ 2 − x = 0, px+ 2 = x, x + 2 = x2, x2− x − 2 = 0, x_1,2= 1± p 1+ 8 2 = −1

2 , only the solutionx= 2 is consistent.

Example 3.11. p1+ x2_{= x + 2, 1 + x}2_{= (x + 2)}2_, _4x_{+ 3 = 0, x = −}₃_/₄_{, the solution is consistent.} Example 3.12. p2x2_{+ 1 = 1 − x, 2x}2_{+ 1 = (x + 2)}2_, _2x2_{+ 1 = 1 − 2x + x}2_, _x2_{+ 2x = 0,} x₁= −2, x₂= 0, both solutions are consistent.

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Precalculus 3.5 Equations with radicals

Example 3.13. p3

x2_{− x − 1 = x − 1,} _x2_{− x − 1 = x}3_{− 3x}2_{+ 3x − 1, x}3_{− 4x}2_{+ 4x = 0, x(x}2₋ 4x+ 4) = 0,

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4 Basic notions of Geometry

The aim of this chapter is to review some fundamental concepts of analytic geometry, also known as coordinate geometry, or Cartesian geometry.

4.1 Cartesian coordinates

We start by considering the Cartesian product R × R × R = R3, i.e. the set of all ordered triples of real numbers. Thanks to the one-to-one correspondence between real numbers and points on a straight line, it is possible to represent the elements of R3as points on a space, which takes the name ofCartesian space. In order to do so, we fix three oriented straights lines, which are called the Cartesian axes and usually are taken to be perpendicular to each other. In latter case, we speak aboutorthogonal Cartesian space. Besides, if all the axes have the same unity of measure the Cartesian space is named mono-metric. In what follows, we always take into accountmono-metric and orthogonal Cartesian spaces. The point (0,0,0) corresponds to the intersection point between the axes (called the origin). In this way, Cartesian axes are indicated byO_x,O_y,O_z, or, simplyx, y, z, and xy, x z, y z are named Cartesian planes.

Instead, if we consider theCartesian plane R × R = R2, the Cartesian axisO_x is calledhorizontal axis or axis of abscissae, whereas O_y vertical axis or axis of ordinates. The Cartesian plane is denoted by O xy and the Cartesian space by O xy z. Once the reference system O xy z is fixed, a one-to-one correspondence is set up between a pointP in the space and an ordered triple of real numbers (the coordinates of the point), as shown in Figure4.1. In order to indicate the coordinates of the pointP we writeP(x, y, z) (P(x, y) into the plane) or, sometimes, P = (x, y, z) (P = (x, y) into the plane).

O x y b P P b xx b yy x y z b P x y z O

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4.2 Fundamental formulae of Geometry

LetA and B two points into the Cartesian space O xy, with respective coordinates(x_A,y_A) and (x_B,y_B). Thedistance between A and B is given by

(4.1) AB=Æ(x_B− xA)2+ (yB− yA)2.

This formula is a direct consequence of the Pythagorean theorem, so the fact that we are working with an orthogonal Cartesian plane is crucial, as shown in Figure4.2below.

O x y A B C b x_A x_A b x_B x_B b y_A y_A b y_B y_B AC = x_C−x A= xB−xA BC = y_B−y C = yB−yA

Figure 4.2 Distance between two points and Pythagorean theorem

Themidpoint M of the line segment joining the points A= (xA,yA) and B = (xB,yB) has Cartesian coordinates(x_M,y_M) given by

(4.2) x_M= xA+ xB

2 , yM =

y_A+ y_B

2 .

Finally, we report the formula for the centroid of a triangle. Precisely, given three verticesA= (x_A,y_A), B= (x_B,y_B) and C = (x_C,y_C) of a triangle ABC , the coordinates of the centroid G are given by

(4.3) x_G= xA+ xB+ xC

3 , yG=

y_A+ y_B+ y_C

3 .

4.3 Lines

The most general equation for a straight line into a Cartesian plane is given by

(4.4) ax+ b y + c = 0

wherea and b represent known numbers. In particular a and b cannot be equal to zero at the same time. In order to draw a line into a Cartesian plane it is sufficient to find out two solutions of Equation (4.4). Example 4.1. Draw into the Cartesian plane the following line: 3x+ 2y − 6 = 0. It is immediate to verify that the two points(0,3) and (2,0) satisfy the equation. The graph is the following:

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Precalculus 4.3 Lines 1 2 3 1 2 3 4 5 −1 −2 −3 b A(0, 3) b B(2, 0) Figure 4.3 Line 3x+ 2y − 6 = 0

Ifb6= 0, Equation (4.4) can be rewritten as

y= −a b x−

c b, or, settingm= −a_b andq= −c_b, as

(4.5) y= mx + q .

The numberm is the line’s slope (or gradient), whereas the number q is called vertical intercept. For example, the line in Figure (4.3) can be written as

y= −3 2x+ 3, withm= −3/₂_and_q= 3. We observe that

(4.6) m= yB− yA

x_B− xA .

The property in previous Equation (4.6) holds for any pair of points of the line. In particular, the numeratory_B− yAand the denominatorxB− xAindicate, respectively, the vertical and the horizontal movement done when moving from the pointA to the point B. The line is: i) increasing, i.e. the slope m is positive, if it goes up from left to right, ii) decreasing, i.e. the slope m is negative, if it goes down from left to right, iii) horizontal, i.e. the slopem is zero, if it is a constant function. The usual formula for the slope is the following one

(4.7) m= ∆y

∆x,

wherethe difference y_B−yAis indicated by∆y (it reads as delta y) and the difference xB−xAis indicated by

∆x (it reads delta x). This is a very important notation and it is commonly used in practice. In particular,

if we consider any variable g , then the difference between two values of g is named variation and it is denoted by∆g. If ∆g is positive (negative) we speak about increment (reduction) of g. For example, if

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the company Alfa has a gain g of 150.000 $ for 2008 and of 180.000 $ for 2009, then∆g = 30.000$. In other words, the company has performed a gain of 30.000 $ over one year.

Vertical lines are characterized by equations of type x= k, i.e. b = 0, while horizontal lines by equations the typey= k, i.e. m = 0. Besides, two non vertical lines are parallel if and only if they have the same slope, whereas they are perpendicular if and only if the slope of the second line is the negative reciprocal of the slope of the first line, i.e. the product between their slopes is−1. In order to find out the equation of a line, we can encounter the following situations.

1. Line passing through a point P and known slope: If P(x_P,y_P) is the point and m is the known slope, than the equation of the line is

(4.8) y− yP= m(x − xP).

2. Line passing through two points A and B: if A(x_A,y_A) and B(x_B,y_B) are two points, then the equation of the line is

(4.9) (x − xA)(yB− yA) = (y − yA)(xB− xA).

Example 4.2. Determine the equation of the line s passing through(1,2) and parallel to the line r : 2x − y+ 5 = 0.

We first have to determine the slope of the line r . To do this, we write it into the form y= 2x − 5: its slope is equal to 2. So, the equation of the lines is y− 2 = 2(x − 1) (or, equivalently 2x − y = 0). Example 4.3. Determine the equation of the line s passing through(2,1) and perpendicular to the line

r : x− 2y − 1 = 0.

We have: r : y=1/₂_x₋1/₂_{. The slope of the line}_{r is} 1

2. Hence, the slope of the lines will be−2 and so the equation of the line isy− 1 = −2(x − 2) (or, equivalently 2x + y − 5 = 0).

Example 4.4. Determine the line passing through(2,3) and (4,−1).

We immediately obtain(x − 2)(−1 − 3) = (y − 3)(4 − 2), which simplifies as 2x + y − 7 = 0.

4.4 Parabolas

4.4.1 Parabola with vertical axis

The equation of a parabola with vertical axis is given by

(4.10) y= ax2+ b x + c, a 6= 0,

wherea, b and c are known numbers, with a6= 0. It has the following fundamental characteristics: – Ifa> 0 its concavity is upward; if a < 0 its concavity is downward.

– The vertexV has abscissa

(4.11) x_V = − b

2a.

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Precalculus 4.4 Parabolas

4.4.2 Parabola with horizontal axis

The equation of a parabola with horizontal axis is given by

(4.12) x= ay2+ b y + c, a 6= 0.

wherea, b and c are known numbers, with a6= 0. It has the following fundamental characteristics: – Ifa> 0 its concavity is rightward; if a < 0 its concavity is leftward.

– The vertexV has ordinate

(4.13) y_V= − b

2a.

– The abscissa ofV can be found by substituting the ordinate y_V into the equation of the parabola. We present now the following examples.

Example 4.5. y = 2x2− x − 1. It has an upward concavity; the vertex has coordinates (1/₄_,₋9/₈). It intersects they axis in(0,−1) and the x-axis in (−1/₂, 0) (these points are obtained by using the quadratic formula for quadratic equations). The graph of the parabola y= 2x2_{− x − 1 is the following:}

1 −1 1 2 −1 bA b B b V

Figure 4.4 Graph of the parabola of equation y= 2x2_{− x − 1}

Example 4.6. x= y2− 2y + 2. It has a rightward concavity; the vertex has coordinates (1, 1). It intersects thex axis in(2,0). To find out the intersections with the y axis we have to put x equal to 0. However, equationy2_{− 2y + 2 = 0 has no solution (the ∆ is negative). We determine other points, (2, 2) and} (5,−1), in order to draw the graph of the parabola:

1 2 −1 1 2 3 4 5 6 7 −1 b V b A b B

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5 Inequalities

A one-variable inequality is an expression of the type

f(x) Ñ g(x) (or, equivalently f (x) < g(x) ∨ f (x) ≤ g(x) ∨ f (x) > g(x) ∨ f (x) ≥ g(x)), while a two-variables inequality reads as

f(x, y) Ñ g(x, y) (or, equivalently f (x, y) < g(x, y)∨f (x, y) ≤ g(x, y)∨f (x, y) < g(x, y)∨f (x, y) ≥ g(x, y)). Solving an inequality means to find a range, or ranges, of values that the variable(s) can take and still satisfy the inequality.

For instance, in the following examples

Example 5.1. – 3x2− 2x > 1: 2 is a solution; 0 is not a solution.

– x2− 2y2≥ x + y: the pairs (2, 0) is a solution; the pair (2, 1) is not a solution.

It is important to note that,usually, in the one-variable inequalities case, there is an infinite number of solutions which can be represented by using subsets of the real numbers (see Chapter2, Paragraph

2.5). Often, it is convenient to represent solutions graphically. This is because it is not always possible to express the set of solutions in an analytical way.

5.1 First order inequalities

5.1.1 First order one-variable inequalities

A first order one-variable inequality assumes one of the following forms

(5.1) ax+ b > 0, ax + b ≥ 0, ax + b < 0, ax + b ≤ 0.

In general, it is convenient to consider the casea> 0, possibly changing the sign of both members. Attention: multiplying or dividing both sides of an inequality by the same negative number the sign of the inequality changes. To solve the inequality, then, one has first to isolate the variable, then to divide b by a. The following examples help to fix ideas.

Example 5.2. 3x+ 2 ≤ 0: 3x ≤ −2, x ≤ −2/₃_{, or} _x_{∈] − ∞, −}2/₃]. This set can also be represented graphically as 0 1 2 0 −1 −2 −3 −4 | −2/₃ b

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Example 5.3. 2x+ 8 < 7x − 1: −5x < −9, 5x > 9, x >9/₅_{, or} _x_∈]9/₅_,+∞[. This set can also be represented graphically as 0 1 2 3 4 5 0 −1 | 9_/₅

Figure 5.2 The inequality 2x+ 8 < 7x − 1

Note 5.1. In Example (5.1), the point−2/₃_{is included into the set of solutions. In the second one, instead,} the point9/₅_{is not included. Usually a (small) filled circle is used to represent the first kind of points (it} is possible to use other conventions. The important thing is to be clear and coherent.).

5.1.2 First order two-variable inequalities

A first order two-variable inequality always assumes one of the following forms (5.2) ax+ b y + c > 0, ax + b y + c ≥ 0, ax + b y + c < 0, ax + b y + c ≤ 0.

Equationax+ b y + c = 0 represents a line into the Cartesian plane. In particular, a line divides the plane in two half-planes. So, a first order two-variables inequality has as solution all the points in one of the two half-planes. The points of the line belong to the set of solutions if the sign= appears into the inequality. In order to know which of the two half-planes has to be selected, it is sufficient to take a point (not belonging to the line) in one of the two half-planes and check, numerically, if this point satisfies the inequality.

Example 5.4. 2x− y + 1 > 0. To solve the inequality the following steps are necessary: i) draw the line 2x− y + 1 = 0, ii) take the point (0, 0) (note that 2 × 0 − 0 + 1 6= 0) and substitute its coordinates into the initial inequality. In particular,(0,0) satisfies the inequality. Than, the set of solutions is given by the half-plane determined by the line 2x− y + 1 = 0 and containing the point (0, 0).

1 2 −1 −2 1 2 3 4 5 −1 −2 −3 bO

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Precalculus 5.2 Inequalities of second order

Example 5.5. 2x+ y + 1 ≥ 0. The set of solutions is represented by the following Figure5.4:

1 2 −1 −2 1 2 3 4 5 −1 −2 −3 b O

Figure 5.4 The inequality 2x+ y + 1 ≥ 0

5.2 Inequalities of second order

5.2.1 One-variable second order inequalities

A second order one-variable inequality has the following representation

(5.3) ax2+ b x + c Ñ 0.

The resolution method is very similar to the first-order case(1). Precisely, one first considers and draws the parabolay= ax2_{+ b x + c, then: i) if the sign of the inequality is ≥ the range of solutions is given} by thexs corresponding to the parts of the parabola which are above the x-axis; ii) if the sign of the inequality is≤ the range of solutions is given by the xs corresponding to the parts of the parabola which are below the x-axis. The following examples will clarify the methodology.

Example 5.6. 2x2− x − 1 ≥ 0. Drawing the parabola 2x2− x − 1 (see Figure5.5), it is evident that the range of solutions of 2x2− x − 1 ≥ 0 are x ≤ −1/₂_or_x_{≥ 1, that is}

x∈ −∞, −1 2 ∪ [1, +∞[.

1_{Actually there is a method that requires the study of the}_{∆ of the quadratic equation ax}2_{+ b x + c = 0, associated to the}

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1 −1 1 2 3 4 −1 −2 −3 b A(1, 0) b B(−1_/₂_{, 0)}

Figure 5.5 The inequality 2x2_{− x − 1 ≥ 0} The range of solutions can be represented, graphically, as follows

0 1 2 3 4 5 0 −1 −2 −3 b 1 b −1/2 b b

Figure 5.6 The range of solutions of the ineuality 2x2_{− x − 1 ≥ 0}

Example 5.7. −2x2_{+ x −1 ≥ 0. Drawing the parabola −2x}2_{+ x −1 (see Figure}_5.7_{), it is evident that the} inequality does not admit any solution. Instead, the inequality−2x2+ x −1 ≤ 0 (resp. −2x2+ x −1 < 0) has as solution the set R (resp. the set R deprived from the element 0).

−1 −2 1 2 3 4 −1 −2 −3 −4

Figure 5.7 The inequality−2x2_{+ x − 1 ≥ 0.}

Example 5.8. x2+ 2x + 1 ≤ 0. Drawing the parabola x2+ 2x + 1 = 0 (see Figure5.8), it is evident that range of solutions of the inequality is given only by the pointx= −1. Indeed, the polynomial x2_{+ 2x + 1 = (x + 1)}2_{is always positive except at}_x_{= −1.}

1

1 2 3

−1 −2 −3

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Precalculus 5.2 Inequalities of second order

5.2.2 Two-variable second-order inequalities

The resolution method is analogous to the case of the two-variable first order inequalities and the following examples will clarify the procedure.

Example 5.9. x2+ y2− 2x − 2y + 1 ≤ 0. To solve this inequality the following steps are necessary: i) draw the circumference of equationx2_{+ y}2_{− 2x − 2y + 1 = 0, ii) check if an internal point (for example} the center) satisfies the inequality, then iii) if the center satisfies the inequality, the set of solutions is given by all the internal points and the points of the circumference (observe that the inequality is of type “≤”). 1 2 1 2 3 −1 b C

Figure 5.9 The inequality x2_{+ y}2_{− 2x − 2y + 1 ≤ 0}

Example 5.10. x2₋y2

4 < 1. Following the same procedure above and trying with the point (0,0), it is possible to conclude that the inequality is verified by all the points between the two branches of the hyperbolax2− y

2

4 = 1. However, in this case the sign of the inequality is <.

1 2 −1 −2 1 2 3 −1 −2 −3

Figure 5.10 The inequality x2₋y2

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5.3 Systems of inequalities

A system of inequalities is a set of inequalities in the same variables. Similarly to the systems of equations case, asolution to a system of inequalities is an assignment of numbers to the variables such that all the inequalities are simultaneously satisfied. In this case, however, the graphical representation will help significantly.

5.3.1 One-variable systems of inequalities

Example 5.11.

2x− 1 ≤ 0

x2− 5x + 4 > 0 . To solve the system it is necessary to solve separately each inequality. The first has as solutionx≤1/₂_{, while the second}_x< 1 ∨ x > 4. Drawing the graph in Figure_5.11_{, it is} easily to deduce that the solutions of the system are given by

−∞,1 2 . 0 1 2 3 4 5 6 7 0 −1 −2 −3 | 1_/₂ | 1 | 4 b

Figure 5.11 Graphical representation of a one-variable system of inequalities

5.3.2 Two-variable systems of inequalities

The method of resolution is very similar to that presented in the previous section. One first solves, separately, each inequality of the system then one intersects the solutions. Also in this case, the graphical representation will be essential. We present the following example

Example 5.12.

x2+ y2− 2x > 0 x− y − 2 > 0 .

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Precalculus 5.4 Factorable polynomial inequalities 1 −1 −2 1 2 3 4 5 −1 −2 −3

Figure 5.12 Graphical representation of a two-variable system of inequalities

The solution set is given by the chequered plane.

5.4 Factorable polynomial inequalities

Let suppose to have a one-variable or a two-variable inequality given by f(x) Ñ 0, f (x, y) Ñ 0.

and that the quantities f(x) or f (x, y) are not a first- or a second-order polynomials. In this case, the so calledrule of signs is used to solve the inequality. Basically the rule of signs states that the product of two numbers with the same signs (resp. different signs) is positive (resp. negative). So, if the polynomial f(x) or f (x, y) is a factorable polynomial, in order to solve the inequality, one has first to determine the sign of each factor, then of the product. To facilitate the analysis, a graphical representation is used. The following examples will clarify the concepts.

Example 5.13. (x2−1)(x −2) > 0. The polynomial f (x) = (x2−1)(x −2) > 0 is a factorable polynomial with factors(x2_{− 1) and (x − 2). In particular, the first factor, x}2_{− 1, is positive for x < −1 and x > 1,} negative for−1 < x < 1, and zero for x = ±1 (y = x2− 1 is the equation of a parabola). The second factor, x− 2 is positive for x > 2, negative for x < 2, and zero for x = 2. The conclusions are easily stated if we consider the graph in the following Figure5.13

0 1 2 3 4 5 0 −1 −2 −3 −4 −5 | −1 | 1 | 2 bc bc bc bc bc bc − + − + x2₋₁ x − 2 Concl.

Figure 5.13 Graphical representation of the sign of the inequality(x2− 1)(x − 2)

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– a plain line is used to indicate the parts where each factor is positive ; – a dotted line is used to indicate the parts where each factor is negative ; – a 0 is used to indicate the points where each factor is exactly equal to zero . Summarising, the inequality is satisfied for

x∈] − 1, 1[ ∪ ]2, +∞[ .

Note that to solve the inequality(x2− 1)(x − 2) < 0 it is not necessary to construct a new graph. In particular,(x2− 1)(x − 2) < 0 is satisfied for

x∈] − ∞, −1[ ∪ ]1, 2[.

Analogously, the inequalities(x2− 1)(x − 2) ≤ 0 and (x2− 1)(x − 2) ≥ 0 has as solutions ] − ∞,−1] ∪ [1,2], and [−1,1] ∪ [2,+∞[, respectively. Example 5.14. x 2_{− 1}

x− 2 ≥ 0. To solve this inequality the rule of signs is again used (the rule of signs is valid also for the quotient between two numbers). The main difference is that, in this case, the factor x− 2 is at the denominator of the fraction and hence, it has to be different from zero. In what follows, the symbol× is used to indicate that, for example, the value x = 2 cannot belong to the set of solutions. Figure5.14represents, graphically, the inequality:

| −1 | 1 × 2 bc bc × bc bc × − + − + x2−1 x − 2 Concl.

Figure 5.14 Graphical representation of the sign of the inequality(x2_{− 1)}_/

(x − 2)≥ 0

The inequality is satisfied for

x∈ [−1, 1] ∪ ]2, +∞[. Example 5.15. x− y + 1

x2+ y2_{− 2y} ≥ 0. The first step consists in determining the sign of the numerator and denominator separately. Then, using the rule of signs one determines the sign of the quotient. In this case, the range of solutions will be a subset of the plane. Figures5.15 and5.16represent the set of positivity of the numerator and of the denominator, respectively. In particular: i) the gray regions indicate the sets of positivity, i) the white regions indicate the sets of negativity, iii) the line and the circle indicate the points in which the numerator and the denominator are equal to zero; these points have to be excluded. Figure5.17represents the set of solutions of the inequality. Note that we have to exclude the entire circumference and the points of intersection between the line and the circumference.

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Precalculus 5.5 Inequalities with radicals 1 2 −1 1 2 −1 −2 Figure 5.15 x− y + 1 > 0 1 2 −1 1 2 −1 −2 Figure 5.16 x2_{+ y}2_{− 2y > 0} 1 2 −1 1 2 3 −1 −2 −3 −4 b c b c Figure 5.17 x− y + 1 x2+ y2_{− 2y} ≥ 0

5.5 Inequalities with radicals

In this course only two types ofinequalities with radicals are considered. In particular: 1. p f(x) ≥ g(x) (or pf (x) > g(x));

2. p f(x) ≤ g(x) (or pf (x) < g(x)).

To solve the first type, one has to consider the union of the solutions of the two systems f(x) ≥ 0 g(x) < 0 ∪ f(x) ≥ g2(x) g(x) ≥ 0 , or f(x) ≥ 0 g(x) < 0 ∪ f(x) > g2(x) g(x) ≥ 0 . To solve the second type, instead, one has to consider the union of the solutions of the two systems:

   f(x) ≥ 0 g(x) ≥ 0 f2(x) ≤ g(x) ,  or    f(x) ≥ 0 g(x) ≥ 0 f2(x) < g(x)  . Example 5.16. px2_{− 9x + 14 > x − 8.} x2− 9x + 14 ≥ 0 x− 8 < 0 ∪ x2− 9x + 14 > (x − 8)2 x− 8 ≥ 0 .

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The first system is satisfied byx≤ 2 ∨ 7 ≤ x < 8; the second system is satisfied by x ≥ 8. So, the range of x values such that x≤ 2 ∨ x ≥ 7 is the solution ofpx2_{− 9x + 14 > x − 8.}

Example 5.17. p4x2_{− 13x + 3 < 2x − 3.}    4x2− 13x + 3 ≥ 0 2x− 3 ≥ 0 4x2− 13x + 3 < (2x − 3)2 . Exercise

To solve a radical inequality containing only a radical with odd index root (in particular, with odd index root equal to 3) it is sufficient to cube both members of the inequality and then, to solve the new inequality.

Example 5.18. p3

x2+ 7 > 2. If we cube both members, we obtain x2_{− 1 > 0, which has as solutions} x< −1 ∨ x > 1.

5.6 Exercises

Exercice 5.1. Solve the following inequalities. 1. x2_{+ 3x + 2 > 0;} 2. −x2− 3x + 2 < 0; 3. 4− x2_{> 0;} 4. x2− x + 6 < 0; 5. (x2_{+ 2x − 8)(x + 1) > 0;} 6. (x2− 2)(x + 1)(1 − x) ≥ 0; 7. x(x2_{+ 2)(2x − 1) < 0;} 8. x+ 1 x2_{+ 1}< 0; 9. 2x− 8 1− x − x2> 0; 10. x 2_{− 4} x+ 3 ≤ 0; 11. x3_{− 27 ≥ 0;} 12. 2− x3< 0; 13. x3_(x2_{− 1)(2 − x}2_{) ≤ 0;} 14. x− 9 x3_{+ 1}≥ 0; 15. 8− x 3 x3_{+ 9}≤ 0.